# Is every finitely dimensional basic $K$-algebra isomorphic to a bound quiver algebra?

Given a positive integer $n$, how to classify $n$-dimensional basic $K$-algebras?, where $K$ is algebraically closed.

For $n=3$, Let $A=\left[ \begin{array}{ccc} K &0& 0\\ 0& K& 0\\ 0 &0& K \\ \end{array} \right] ,B=\left[ \begin{array}{cc} K & 0 \\ K & K\\ \end{array} \right]$ and $C=\left[ \begin{array}{ccc} K &0& 0\\ K&0& 0\\ K &0& 0\\ \end{array} \right]$. Then we have $\operatorname{rad}A=\left[ \begin{array}{ccc} 0 &0& 0\\ 0& 0& 0\\ 0 &0& 0 \\ \end{array} \right] ,\operatorname{rad}B=\left[ \begin{array}{cc} 0 & 0 \\ K & 0\\ \end{array} \right]$ and $\operatorname{rad}C=\left[ \begin{array}{ccc} 0 &0& 0\\ K&0& 0\\ K &0& 0\\ \end{array} \right]$, and hence $A/\operatorname{rad}A\cong K\times K\times K, B/\operatorname{rad}B\cong K\times K, C/\operatorname{rad}C\cong K$, this implies that $A,B$ and $C$ are basic three-dimensional algebras. Let $Q$ is the quiver

$$\circlearrowright^{\beta}$$ and $\mathcal{I}$ is the ideal of$KQ$ generated by one zero relation $\beta^3$. Then $D=KQ/I$ is basic, Are there other three dimensional basic algebra which are not isormorphic to the above? It is known that every bound quver algebra is basic, conversely, Is every $n$-dimensional basic $K$-algebra isomorphic to a bound quiver algebra?

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en.wikipedia.org/wiki/… –  Jackson Walters Nov 29 '12 at 2:19
In principle you could write down all quivers with number of vertices + number of arrows $\leq n$. Then try to write down all relations making it an $n$-dimensional algebra. This could be done with some effort, as long as $n$ is not too big. But the reallly hard task is to decide whether two such algebras are isomorphic. I doubt there is an easy receipe unless $n$ is really small, maybe $\leq 10$. –  Julian Kuelshammer Nov 29 '12 at 8:41
What is a “basic” algebra? –  Ewan Delanoy Dec 3 '12 at 8:39
Assume that $A$ is a $K$-algebra with a complete set ${e_1,\cdots , e_n}$ of primitive orthogonal idempotents. The algebra $A$ is called basic if $e_iA \ncong e_jA$, for all $i\neq j$. –  Aimin Xu Dec 3 '12 at 10:09
@AiminXu There is one more $3$-dimensional algebra: There are two algebras with $C/\operatorname{rad} C\cong K$, one has $\operatorname{rad}^2 C=0$, another one has $\operatorname{rad}^2 C\cong \operatorname{rad} C\cong K$. –  Julian Kuelshammer Dec 4 '12 at 10:13

Let $A$ be a basic and connected finite dimensional Kalgebra, then it is isomorphic to a bound quiver algebra by Theorem 3.7 p64, ( I. Assem, D. Simson and A. Skowro$\acute{n}$ski, Elements of the Representation Theory of Associative Algebras, Volume 1) –  Aimin Xu Dec 9 '12 at 3:13
Is every finite dimensional basic $K$-algebra connected? –  Aimin Xu Dec 9 '12 at 3:24